Trader Intern Interview Questions | Glassdoor

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Trader intern interview questions shared by candidates

## Top Interview Questions

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Jan 12, 2011
 Russian Roulette - 4 blanks 2 bullets, all in a row. If someone shoots a blank next to you, would you take another shot or spin 12 Answers take another shot, 3/4 chance of surviving vs 2/3 if you respin Here is my answer: the prob. of survival after re-spin is: 3/5. the prob. of survival with on re-spin is: 1/15 * 6/4 = 1/10. correction: the prob. of survival after re-spin is: 4/6. Show More Responses my final correction: the prob. of survival after re-spin is: 4/6 = 2/3 the spin with no spin is: 2/5 = c(4, 2) / c(6, 2). Denoting the blanks by 0's and live bullets by 1's, and adjoining the left and right edges (denoted by ~) , so as to make a cylinder, the following diagram illustrates how the bullets are arranged in the cylinder of the revolver: ~000011~ . Denote the bullet that is to be fired if the trigger is pulled by enclosing it in parenthesis. Denote an empty chamber by *. Assume that when you pull the trigger, the cylinder rotates clockwise (to the right in the diagram above). If when you pulled the trigger for the first time, the bullet was a blank, then before and after you pulled the trigger, the cylinder was and is in one of the following states: 1) Before: ~(0)00011~ After: ~*0001(1)~ 2) Before: ~0(0)0011~ After: ~(0)*0011~ 3) Before: ~00(0)011~ After: ~0(0)*011~ 4) Before: ~000(0)11~. After: ~00(0)*11~ If you pull the trigger again without spinning the cylinder, then only in case one will the bullet be a live bullet, yielding a 3/4 probability of the bullet being a blank . If you spin the cylinder and then pull the trigger, then you have a 4/6=2/3 probability of the bullet being a blank. Clearly, you should not spin the cylinder. don't spin, 3/4 prob survival if not spinned respin - 4/6 = 2/3 (obvious) don't spin - you only die if the blank was the "last" one, which is 1/4 chance. hence 3/4 chance of live. since 3/4>2/3 don't spin. Wouldn't play. Spin Spin: 4/6=0.6667 to survive Not Spin: C(4,2)/C(5,2)=3/5=0.6 condition on 2 bullet is in consecutive position and someone survives the 1st shot, we have: spin it-> survival rate is 4/6=2/3=67% not spin it-> survival probability=3/4=75% so...not spin it will have a bigger chance of survival. the question did not say that the bullets are next to each other so in this case you should spin I. Using Bayes' Theorem: P[0] = probability of a blank : 4/6 P[1] = probability of a bullet : 2/6 P[1|0] = probability of a bullet given a blank : find this P[0|1] = probability of a blank given a bullet If a bullet occurs then the next pull can be a bullet or a blank and so P[0|1] = 1/2 P[1|0] = P[0|1]*P[1] / P[0] = 1/4 So there is a 25% chance that the next pull is a bullet, or a 75% chance that it is a blank. The first pull had a probability of 8/12 of being a blank. Given a blank on the first pull, the second pull has a 9/12 probability of being a blank. II. From a frequentist perspective: So when the first pull is made and it is a blank the event space decreases from 6 possible outcomes to 4 possible outcomes. The first pull was on the last bullet or the the first pull was on the second to last blank. Out of the four possible events there is one where after the pull the chamber is sitting on the blank right before the bullet. Out of the four events 3 are blanks (3/4) and 1 is a bullet (1/4).

Sep 20, 2011
 What if you could reflip 1 coin that you wanted. What would be the expected value then? 6 Answers So I flip all my coins for the first time and my expected value is 2. Now, if all my coins are already heads (1/16 of the time) I stop and get \$4 (1/16*4) = .25 If I didn't get all heads the first time I select any one of the tails and re-flip, and then have a 50% chance of increasing my payout by \$1, or effectively, I theoretically increase my payout by (.5)(1) = .50 cents. So the final EV is: .25 + (15/16)(2.5) = 2.59375 Not Quite prob dist (1/16,1/4,6/16,1/4,1/16) for (4h,3h,2h,1h,0h,), payoff terminal node (4,4,3,2,1), EV(Risk Neutral)=41/16 Show More Responses it's 79/32 79/32 is correct the post before was wrong to assume that a refip gets you a guarenteed heads usual expectation is 2, now we have to evaluate this option of flipping one additional coin. 1/16 of the time it's worthless, 15/16 of the time it's (conditional) value is 1/2. so the answer is 2 + 15/32 = 79/32

Oct 12, 2010
 What is the smallest number divisible by 225 that consists of all 1s and 0s? 7 Answers 11111111100 Two facts to note are: 1) Multiplies of 225 end in 00 25 50 75 2) A number is divisible by 9 if the digits sum to 9. Only those multiples ending in 00 could have only 1's and 0's. So, the smallest digit we can multiply 225 by to get 00 at the end is 4. That is, 225*4=900. Now, we want to find the smallest multiple of 900 that contains only 1's and 0's. Let us first focus on 9 and then tack on the 00 after. The smallest multple of 9 with only 1's and 0's is 111111111. This is a consquence of the fact that the digits must sum to 9. Now, we tack on the 00 at the end and obtain 11111111100. ^^^^^^WRONG 1111111110. Think of 225 as (5)^2(9). So this number must end in a 0 to be divisible by 5. Since every number that 5 divides ends in a 5 or 0, that number is also divisible by 5. Since a number divisible by 9 must have the digits sum to a number divisible by 9, then all we need is 9 1's and a 0 on the end for 1111111110. Show More Responses 1111111110 / 225 = 4938271.6 I fail to understand the above answers. Shouldn't it be 225 or 2250. 225/225 =1 and 2250/225 =10. Both these numbers contain just 1's and 0's :P The above answers seem too complicated/not explained at all. First, notice that 225 = 25 * 9. (1) A number is divisible by 9 iff the sum of its digits is divisible by 9 => we must have 9 1's in our number. (2) A number is divisible by 25 iff the last two digits are divisible by 25 => our last two digits must be 0's. Putting (1) and (2) together, our number is 11111111100. 1000-100

Mar 2, 2010
 What is the chance that at least two people were born on the same day of the week if there are 3 people in the room? 6 Answers ~89.8% incorrect. 1-364/365*363/365 = 1-.9918 = .0082...something like this for 2 people, for 2+ add in the chance of 3 of 3. So first guy then gets 365/365 choices, next two get 1/365 Also incorrect... They are asking if they were born the same day of the week not the exact same day as Chase alluded to. In this case the easiest way to do the problem is 1- probability that no one was born on the same day. 1-(1)(6/7)(5/7)+ 19/49 or 38.78% Show More Responses My bad =19/49 Or you can calculate it directly: The chance of SELECTED two people(but not three) that are born on the same day is 1* 1/7 - 1/7*1/7 =6/49 Now we have 3 chooses 2 = 3 possible pairs, so the probability that any PAIR(but not all three) that are born on the same day is 3* 6/49 = 18/49. However, since the question asks "at least 2", we will need to add the possibility that all three are born at the same day, which is 1/7 * 1/7 = 1/49. So the final result is 18/49 + 1/49 = 19/49 This obviously is a worse way of calculating it, but it's an alternative for some people who wants to think "straight". 1-(7/7*6/7*5/7) = 19/49 = 0.39 # 1 minus the opportunities that all 3 people were born in different day of the week.

Jan 29, 2012

Sep 20, 2011
 You have 4 coins. You throw them in the air. For every H that lends, you get 1 dollar. How much would you pay to play this game? 6 Answers 1/16(0)+4/16(1)+6/16(2)+4/16(4)+1/16(4) = \$2 Sorry; 1/16(0) + 4/16(1) + 6/16(2) + 4/16(3) + 1/16(4) = \$2 Or just treat each coin as a separate event. 50% chance of flipping a H. .5 x \$1 = \$0.50 expected value per coin. \$2 total. Show More Responses binomial distribution, the expectaion value is just N*P so 4*0.5. Working out each ontirbution wouldprobably take longer than what they're looking for. should be less than \$2 why would you paying to play a game just to break even. Isn't common sense 2?

Dec 12, 2011
 question about having 12 marble (1 weigh different) , and one balance. Ask to find the different weight one with least try. 3 Answers Case 1. 5 and 5 either side. if they are equal, check the 2 left over. (2 times) Case 2 5 and 5 either side. if not equal check the heavier side, chose 4 marbles 2 either side of scales. if they are equal its the one left out if they are not use scales again. 3 times This is really a complicate question. We should remember that we don't know the different marble is heavier or lighter. The correct answer should be: Separate 12 marbles into 3 groups, each one has 4 marbles. Mark the 12 marbles as ABCD, EFGH, IJKL. First weight ABCD and EFGH. Case one: If ABCD=EFGH, different one is in I,J,K,L. Second weight AIJK and EFGH If AIJK=EFGH, L is different, then the third time weight L and any other one, and know everything. If AIJK > EFGH, the different one is heavier and is in I,J,K. Third time weight I and J If I> J, I is the different; if I J, J is the different; if I EFGH, the different one is in A, B, C, D, E, F, G, H, and I=J=K=L Second time weight AEFI and GHJK If AEFI = GHJK, the different one is in B, C, D, and heavier. Third time weight B and C. If AEFI>GHJK, the different one is in A, E, F, G, H. Because ABCD>EFGH and AEFI>GHJK, Only A is heavier or one of G, H is lighter. Third time weight G and H. If G>H, H is different one and lighter; If GEFGH and AEFI

Sep 20, 2011
 Now, you have an option to reflip all the coins if you want. What's your strategy? And how what's the expected value? 5 Answers 2.375 To elaborate: Flip again if 0,1, or 2 heads. If 3 or 4 heads, (5/16 of the time) EV = (1/16(4) + 1/4(3)) = \$1 If 0, 1, 2; (11/16 of the time), reflip, and EV = \$2 So, assuming we "choose" to flip again, EV = \$1 + (11/16)(\$2) = \$2.375 as an addendum to the above, it makes no difference to hte expectation value if decide you flip all again when you got the expectation value the first time. i.e. the result is the same if you decide to only roll again if you get a 0 or a 1. Show More Responses The question asks the expected value of the new flip, we should only flip again if the first run was below expected value, and so we should flip all 4 coins which the expected pay off is \$2 dollar again. 2.48